Let the time to fill the conical flask be x minutes. 
In 1 minute 20 m water flows.
So in x minutes 20x m will flow.
\(\therefore\)  Volume of water flowed through cylindrical pipe
= \(\pi r^2\) (20x) x 100 cm
= \(\pi\) (2)² (20x) x 100 cm
Volume of water filled in conical tank
= \(\frac{1}{3}\pi\) (40)² x 72 cm
Volume of water through pipe = Volume of water filled in conical tank
\(\Rightarrow\)  \(\pi\)(2)² (20x) 100 = \(\frac{\pi}{3}\) (40)² x 72
\(\Rightarrow\) x = \(\frac{24}{5}\) minutes = 4\(\frac{4}{5}\) minutes
Hence, conical tank filled in 4\(\frac{4}{5}\) minutes

We have,
Volume of water that flows per hour
   = (192.50 x 60) litres
   = (192.50 x 60 x 1000) cm³     ........(i)
Inner diameter of the pipe = 7 cm
Inner radius of the pipe \(\frac{7}{2}\) cm = 3.5 cm
Let h cm be the length of the column of water that flows in one hour.
Clearly, water column forms a cylinder of radius 3.5 cm and length h cm.
Volume of water that flows in one hour = Volume of the cylinder of radius 3.5 cm and length h cm.

Radius of hemisphere portion = Radius of cylindrical portion = \(\frac{7}{2}m\)
Total length of the tanker = 20 m
\(\therefore\) Length of cylindrical portion = 20 - 7 = 13 m

Now, Capacity (volume) of the oil tanker
= Volume of cylinder + 2 x Volume of hemisphere
= \(\pi r^2h+2\times\frac{2}{3}\pi r^3\)
= \(\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times13+2\times\frac{2}{3}\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times\frac{7}{2}\)
= 500.5 + 179.67 = 680.17 m³

Radius of base of each of the cone and the cylinder (r) = 8 cm
Height of cylindrical part (h₁) = 240 cm
Heigth of conical part (h₂) = 36 cm
Colume of the iron pillar = Volume of cylinder + Volume of cone

  = \(\pi (8)^2\times240+\frac{1}{3}\pi (8)^2\times36\)
  = 64\(\pi\) (240 + 12)
  = 64 x \(\frac{22}{7}\) x 252 = 50688 cm³
\(\therefore\) Weight of the piller = 7.5 x 50688 g
= 380160g = 380.16 kg.

Here, radius of cone and cylinder = 6 cm
Height of cone and cylinder = 7 cm
Slant height of cone = \(\sqrt { { \left( 6 \right) }^{ 2 }+{ \left( 7 \right) }^{ 2 } } \)
= \(\sqrt{36+46}\)
= \(\sqrt{85}\)
= 9.22 cm
 
Now, total surface area of the remaining solid = C.S.A. of cylinder + C.S.A. of cone + area of base of cylinder= \(2\pi rh+\pi rl+\pi r^2\)
= \(\pi r\)(2h + l + r)
= \(\frac{22}{7}\) x 6(2 x 7 + 9.22 + 6)
= \(\frac{22}{7}\) x 6 x 29.22 = 551 cm²

Here, radius of light circular cone (r) = 3cm.
Let h and l be the height and slant height of the right circular cone respectively.
Now, \(\pi rl\) = 47.1 cm²                [given]
\(\Rightarrow\)     3.14 x 3 x l = 47.1
\(\Rightarrow\)     l = \(\frac{47.1}{3.14\times3}\) = 5 cm
Also,   r² + h² = l²
\(\Rightarrow\)     3² + h² = 5²
\(\Rightarrow\)     h² = 25 - 9 = 16
\(\Rightarrow\)     h = 4 cm
\(\therefore\) Volume of the cone = \(\frac{1}{3}\pi r^2h\)
= \(\frac{1}{3}\) x 3.14 x 3 x 3 x 4
= 37.68 cm³ 

Let radius of the upper face = BE = x
\(\triangle\)ABE ~ \(\triangle\)ACD
\(\Rightarrow\)   \({{AB}\over{AC}}={{BE}\over{CD}}\)
\(\Rightarrow\)  \({{4}\over{12}}={{x}\over{6}}\Rightarrow x=2\) cm

Remaining solid is a frustum with height
= 12 - 4 = 8 cm
r₁ - 2 cm, r₂ = 6 cm
Let Slant height be 
l = \(\sqrt{{h}^{2}+{({r}_{2}-{r}_{1})}^{2}}\)
l = \(\sqrt{64+16}\)
l = \(\sqrt{80}\)
l = \(4\sqrt 5\) cm
Now, total surface area
= \(\pi l({r}_{1}+{r}_{2})+{ \pi r }_{ 1 }^{ 2 }+{ \pi r }_{ 2 }^{ 2 }\)
= \(\pi[l({r}_{1}+{r}_{2})+{ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }]\)
= \({{22}\over{7}}[4\sqrt{5}(2+6)+4+36]\)
= \({{22}\over{7}}\times[32\sqrt{5}+40]\)
= \({{22}\over{7}}(32\times2.236+40]\)
= 350.592 cm²

Let height of the cone be x cm
Radius of the cone = 3.5 cm = radius of hemisphere

Volume of the toy = \({{1}\over{3}}{\pi r}^{2}h+{{2}\over{3}}{\pi r}^{3}\)
\(166{{5}\over{6}}={{1}\over{3}}\times{{22}\over{7}}{(3.5)}^{2}\times x+{{2}\over{3}}\times{{22}\over{7}}\times{(3.5)}^{3}\)
\({{1001}\over{6}}={{269.5}\over{21}}x+{{1886.5}\over{21}}\)
\(\Rightarrow\)   \({{1001}\over{6}}-{{1886.5}\over{21}}={{269.5}\over{21}}x\)
\(\Rightarrow\)    \({{3234\times21}\over{42\times269.5}}=x\Rightarrow 6\) cm²
\(\therefore\)   Height of toy = 6 cm + 3.5 cm = 9.5 cm
Surface area of hemicspherical part = \({2\pi r}^{2}\)
= \(2\times{{22}\over{7}}\times3.5\times3.5=77\) cm²
Cost to paint at a rate of Rs 10 per cm²
= 77 x 10 = Rs 770

Consider an area of cross-section of canal as ABCD.
Area of cross-section = 6 × 1.5 = 9 m²
Speed of water = 10 km/h = 10000/60 metre/min
Volume of water that flows in 1 minute from canal = 9 x 10000/60 =1500 m³
Volume of water that flows in 30 minutes from canal = 30 × 1500 = 45000 m³
Let the irrigated area be A. Volume of water irrigating the required area will be equal to the volume of water that flowed in 30 minutes from the canal.
Vol. of water flowing in 30 minutes from canal = Vol. of water irrigating the reqd. area
4500 = (Ax8)/100
A = 562500 m²
Therefore, area irrigated in 30 minutes is 562500 m².

Volume of kerosene in case
= 20 x 20 x 45 = 25 x 15 x h (Volume in second cane)
\(\Rightarrow\)  h = \(\frac{20\times20\times45}{25\times15}\) = 48 cm

Suppose in x hours water will be 3 metres deep in the tank.
Volume of water in the tank = (150 x 100 x 3) m³ = 45000 m³
Area of cross-section of the pipe = \((\frac{2}{10}\times\frac{1.5}{10})m^2=\frac{3}{100}m^2\)
Volume of water that flows in the tank in x hours
= (area of cross-section of the pipe) x (speed of water) x (time)
= \((\frac{3}{100}\times15000\times x)\) m³ = 450 x m³
[\(\therefore\) Speed = 15 km/h = 15000 m/h]
Since the volume of water in the tank is equal to the volume of water that flows in the tank in x hours.
\(\therefore\)  450 x = 45000   \(\Rightarrow\) x = 100 hours.

R= 20cm, r= 10 cm h=21 cm
Capacity of bucket = \(\frac { 1 }{ 3 } \pi h\left( R^{ 2 }+r^{ 2 }+Rr \right) \)
= \(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 21\times \left( 20 \right) ^{ 2 }+\left( 10 \right) ^{ 2 }+20\times 10\)
= 22 (400 +100 +  200)
= 22 x 700 = 15400 cm³ = \(\frac { 15400 }{ 1000 } l=15.4l\)
Total cost of milk at the rate of Rs 30 per litre = Rs 30 X 15.4 = Rs 462.00

Radii of bucket are 14 cm and r cm
h = 15cm
Volume of the bucket = \(\frac { 1 }{ 3 } \pi h(R^{ 2 }+R\times r+r^{ 2 })\)
\(\Rightarrow 5390=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 15(196+14r+r^{ 2 })\)
\(\Rightarrow \frac { 5390\times 7 }{ 22\times 5 } =196+14r+r^{ 2 }\)
\(\Rightarrow 343-196=14r^{ 2 }+r^{ 2 }\Rightarrow r^{ 2 }+14r-147=0\)
\(\Rightarrow r^{ 2 }+21r-7r-147=0\Rightarrow r(r+21)-7(r+21)=0\)
\(\Rightarrow (r-7)(r+2)=0\Rightarrow r=7,-21\) (Rejected)
\(\therefore \) r= 7cm

Length of reservoir = 120m and width = 75m
Height of water = 2.4 m
Volume of water flow in 18 hrs = 120 x 75 x 2.4 = 21600 m³
Hence volume of water that should flow in 1 hr = \(\frac { 21600 }{ 18 } =1200\quad m^{ 2 }\)
Area of cross-section of pipe = \(\frac { 20 }{ 100 } \times \frac { 20 }{ 100 } =.04m^{ 2 }\)
\(\therefore \) Length of water column in 1 hour = \(\frac { volume }{ area } =\frac { 1200 }{ 04 } =30000m\)
Speed of water = 30000 m/h = 30 km/h

Thickness of the log = \(\frac{volume}{l\times b}=\frac{1.104}{2.3\times0.75}\)m = 0.64 m
Number of planks = \(\frac{thickness \ \ of \ \ the \ \ log}{thickness \ \ of \ \ one \ \ plank}=\frac{0.64}{0.04}\) = 16